General volumes and Minkowski problem for G(t, ·) decreasing
4.3 Minkowski-type problems
d eCG([h1], u),
where hε is given by (4.29) with hK and hL replaced by h1 and h2, respectively.
Again, the following corollary is a direct consequence of the previous theorem with G = Φ or Φ.
Corollary 4.2.9. Let h1, h2 ∈ C+(Ω) and let ϕ1, ϕ2 ∈ I or ϕ1, ϕ2 ∈ D. Suppose that for i = 1, 2, ϕi is continuously differentiable and such that ϕ′i is nonzero on (0, ∞).
If φ : Rn\ {o} → (0, ∞) and Φ (or Φ, as appropriate) are continuous, then
n ϕ′1(1)
Z
Ω
ϕ2
h2(u) h1(u)
d eCφ,V([h1], u) =
ε→0lim+
Vφ([h1]) − Vφ([hε]) ε
ε→0lim+
Vφ([hε]) − Vφ([h1])
ε ,
where hε is given by (4.29) with hK and hL replaced by h1 and h2, respectively.
4.3 Minkowski-type problems
This section is dedicated to providing a partial solution (Gt < 0) to the Orlicz-Minkowski problem for the measure eCG,ψ(K, ·).
Proposition 4.3.1. Let G and Gt be continuous on (0, ∞) × Sn−1, let ψ : (0, ∞) → (0, ∞) be continuous, and let K ∈ K(o)n. The following statements hold.
(i) The signed measure eCG,ψ(K, ·) is absolutely continuous with respect to S(K, ·).
(ii) If Ki ∈ K(o)n, i ∈ N, and Ki → K ∈ K(o)n as i → ∞, then eCG,ψ(Ki, ·) → eCG,ψ(K, ·) weakly.
(iii) If Gt > 0 on (0, ∞) × Sn−1 (or Gt < 0 on (0, ∞) × Sn−1), then eCG,ψ(K, ·) (or
− eCG,ψ(K, ·), respectively) is a nonzero finite Borel measure not concentrated on any closed hemisphere.
Proof. (i) Let E ⊂ Sn−1 be a Borel set such that S(K, E) = 0. If g = 1E, the left-hand side of (4.8) is eCG,ψ(K, E). This equals the expression in (4.9), in which we observe that since K ∈ K(o)n, for x ∈ ∂K both |x| and hx, νK(x)i = hK(νK(x)) are bounded away from zero and bounded above, and hence our assumptions imply that
sup
x∈∂K
ρK(¯x) Gt(ρK(¯x), ¯x) hx, νK(x)i ψ(hx, νK(x)i) |x|n
= c < ∞,
where ¯x = x/|x|. Then from (4.8) and (4.9) we conclude, using (2.5), that
eCG,ψ(K, E) ≤ c Z
∂K
1E(νK(x)) dx = c Hn−1(νK−1(E)) = c S(K, E) = 0.
(ii) Let g : Sn−1 → R be continuous and let
IK(u) = g(αK(u))ρK(u) Gt(ρK(u), u) ψ(hK(αK(u)))
be the integrand of the right-hand side of (4.8). Suppose that Ki ∈ K(o)n, i ∈ N, and Ki → K ∈ K(o)n. By [29, Lemma 2.2], αKi → αK and hence, by the continu-ity of Gt and the continuity of the map (K, u) 7→ hK(u) (see [59, Lemma 1.8.12]), IKi → IK, Hn−1-almost everywhere on Sn−1. Moreover, our assumptions clearly yield sup{IKi(u) : i ∈ N, u ∈ Sn−1} < ∞. It follows from (4.8) and the dominated convergence theorem that
Z
Sn−1
g(u) d eCG,ψ(Ki, u) → Z
Sn−1
g(u) d eCG,ψ(K, u) as i → ∞, as required.
(iii) Suppose that Gt> 0 on (0, ∞) × Sn−1; the case when Gt< 0 on (0, ∞) × Sn−1 is similar. Let m = minx∈∂KJK(x), where
JK(x) = ρK(¯x) Gt(ρK(¯x), ¯x) hx, νK(x)i
ψ(hx, νK(x)i) |x|n , x ∈ ∂K,
and ¯x = x/|x|. Since K ∈ K(o)n, our assumptions imply that m > 0. By (4.8) and (4.9),
Z
Sn−1
hu, vi+d eCG,ψ(K, v) = Z
∂K
hu, νK(x)i+JK(x) dx
≥ m Z
∂K
hu, νK(x)i+dx
= m Z
Sn−1
hu, vi+dS(K, v)
> 0.
This shows that eCG,ψ(K, ·) also satisfies (2.6).
In view of Proposition 4.3.1 (iii), one can ask the following Minkowski-type problem for the signed measure eCG,ψ(·, ·).
Problem 4.3.2. For which nonzero finite Borel measures µ on Sn−1 and continuous functions G : (0, ∞) × Sn−1 → (0, ∞) and ψ : (0, ∞) → (0, ∞) do there exist τ ∈ R and K ∈ K(o)n such that µ = τ eCG,ψ(K, ·)?
It follows immediately from (4.10), on using [54, (2.2), p. 93 and (3.28), p. 106], that solving Problem 4.3.2 requires finding an h : Sn−1→ (0, ∞) and τ ∈ R that solve (in the weak sense) the Monge-Amp`ere equation
τ h
ψ ◦ hP ( ¯∇h + hι) det( ¯∇2h + hI) = f, (4.51) where P (x) = |x|1−nGt(|x|, ¯x) for x ∈ Rn. Here f plays the role of the density function of the measure µ in Problem 4.3.2 if µ is absolutely continuous with respect to spherical Lebesgue measure. Formally, then, Problem 4.3.2 is more difficult, since it calls for h in (4.51) to be the support function of a convex body and also a solution for measures that may not have a density function f . The Minkowski problem in [54, Problem 8.1] requires finding, for given p, q ∈ R, n-dimensional Banach norm k · k, and f : Sn−1 → [0, ∞), an h : Sn−1→ (0, ∞) that solves the Monge-Amp`ere equation h1−pk ¯∇h + hιkq−ndet( ¯∇2h + hI) = f (4.52) on the unit sphere Sn−1, where ¯∇ and ¯∇2 are the gradient vector and Hessian matrix of h, respectively, with respect to an orthonormal frame on Sn−1, ι is the identity map on Sn−1, and I is the identity matrix.
To see that (4.51) is more general than (4.52), note firstly that the homogeneity
of the left-hand side of (4.52) allows us to set τ = 1, without loss of generality (if p 6= q, which is true in the case p > 0, q < 0 of particular interest in the present chapter). Let p, q ∈ R and let Q ∈ Sc+n. For t > 0 and u ∈ Sn−1, we set ψ(t) = tp and G(t, u) = (1/q)tqρQ(u)n−q, if q 6= 0, and G(t, u) = (log t)ρQ(u)n, otherwise. (When q ≤ 0, we have G : (0, ∞) × Sn−1 → R and Remark 4.2.6 applies.) Then, using the fact that ρQ is homogeneous of degree −1, we have P (x) = ρQ(x)n−q, for q ∈ R and x ∈ Rn\ {o}. Therefore (4.51) becomes
h1−pk ¯∇h + hιkq−nQ det( ¯∇2h + hI) = f,
where k · kQ = 1/ρQ is the gauge function of Q. Note that k · kQ is an n-dimensional Banach norm if Q is convex and origin symmetric.
Our contribution to Problem 4.3.2 is as follows. Recall that Σε(v) = {u ∈ Sn−1 : hu, vi ≥ ε} for v ∈ Sn−1 and ε ∈ (0, 1).
Theorem 4.3.3. Let µ be a nonzero finite Borel measure on Sn−1 not concentrated on any closed hemisphere. Let G and Gt be continuous on (0, ∞) × Sn−1 and let Gt < 0 on (0, ∞) × Sn−1. Let 0 < ε0 < 1 and suppose that for v ∈ Sn−1,
t→0+lim Z
Σε0(v)
G(t, u) du = ∞ and lim
t→∞
Z
Sn−1
G(t, u) du = 0. (4.53)
Let ψ : (0, ∞) → (0, ∞) be continuous and satisfy Z ∞
1
ψ(s)
s ds = ∞. (4.54)
Then there exists K ∈ K(o)n such that µ
|µ| = CeG,ψ(K, ·)
CeG,ψ(K, Sn−1). (4.55)
Proof. Note that the limits in (4.53) exist, since t 7→ G(t, u) is decreasing. Define
ϕ(t) = Z t
1
ψ(s)
s ds, t > 0, (4.56)
and
a = − Z 1
0
ψ(s)
s ds ∈ R ∪ {−∞}. (4.57)
Then, by (4.54), (4.56), and (4.57), ϕ ∈ Ja is strictly increasing and continuously differentiable with tϕ′(t) = ψ(t) for t > 0; the latter equality implies that ϕ′is nonzero on (0, ∞). For f ∈ C+(Sn−1), let
F (f ) = 1
|µ|
Z
Sn−1
ϕ(f (u)) dµ(u), (4.58)
and for K ∈ K(o)n, define F (K) = F (hK). We claim that
α = infn
F (K) : eVG(K) = |µ| and K ∈ K(o)n
o
(4.59) is well defined with α ∈ R ∪ {−∞} because there is a K ∈ K(o)n with eVG(K) = |µ|.
To see this, note that
VeG(rBn) = Z
Sn−1
G(r, u) du ≥ Z
Σε0(v)
G(r, u) du
for any v ∈ Sn−1. Then (4.53) yields eVG(rBn) → ∞ as r → 0, and eVG(rBn) → 0 as r → ∞. Since r 7→ eVG(rBn) is continuous, there is an r0 > 0 such that eVG(r0Bn) = |µ|.
It follows from (4.59) that α ∈ R ∪ {−∞}.
By (4.59), there are Ki ∈ K(o)n, i ∈ N, such that eVG(Ki) = |µ| and
i→∞lim F (Ki) = α. (4.60)
We aim to show that there is a K0 ∈ K(o)n with eVG(K0) = |µ| and F (K0) = α.
To this end, we first claim that there is an R > 0 such that Ki∗ ⊂ RBn, i ∈ N. Suppose on the contrary that supi∈NRi = ∞, where Ri = maxu∈Sn−1ρKi∗(u) = ρKi∗(vi). By taking a subsequence, if necessary, we may suppose that vi → v0 ∈ Sn−1 and limi→∞Ri = ∞. There exists i0 ∈ N such that |vi− v0| < ε0/2 whenever i ≥ i0. Hence, if u ∈ Σε0(v0) and i ≥ i0, then hu, vii ≥ ε0/2. It follows that for u ∈ Σε0(v0) and i ≥ i0, we have
hKi∗(u) ≥ ρK∗i(vi)hu, vii
= Rihu, vii
≥ Riε0/2,
and therefore
as i → ∞. This contradiction proves our claim.
By the Blaschke selection theorem, we may assume that Ki∗ → L for some L ∈ K n. Suppose that L /∈ K(o)n. Then o ∈ ∂L, so there exists w0 ∈ Sn−1 such that uniformly on Sn−1. The continuity of ϕ ensures that
sup{|ϕ(hKi(u))| : i ∈ N, u ∈ Sn−1} < ∞.
Now it follows from (4.58), (4.60), and the dominated convergence theorem that
α = lim
i→∞F (Ki) = 1
|µ|
Z
Sn−1
i→∞lim ϕ(hKi(u)) dµ(u) = 1
|µ|
Z
Sn−1
ϕ(hK0(u)) dµ(u) = F (K0).
(4.61) Also, by Lemma 4.1.2, we have eVG(K0) = |µ|, so the aim stated earlier has been achieved. It also follows from (4.61) that α ∈ R.
We now show that K0 satisfies (4.55) with K replaced by K0. Due to ϕ ∈ Ja and f ≥ h[f ], one has F (f ) ≥ F (h[f ]) = F ([f ]) for f ∈ C+(Sn−1). By (4.61),
F (hK0) = F (K0) = α = inf{F (f ) : eVG([f ]) = |µ| and f ∈ C+(Sn−1)}. (4.62) Let g ∈ C(Sn−1). For u ∈ Sn−1 and sufficiently small ε1, ε2 ≥ 0, let hε1,ε2 be defined by (4.28) with f0 and εg replaced by hK0 and ε1g + ε2, respectively, i.e.,
hε1,ε2(u) = ϕ−1(ϕ(hK0(u)) + ε1g(u) + ε2) . (4.63) Then for sufficiently small ε, we have
hε1+ε,ε2(u) = ϕ−1(ϕ(hε1,ε2(u)) + εg(u)) , and
hε1,ε2+ε(u) = ϕ−1(ϕ(hε1,ε2(u)) + ε) .
The properties of ϕ listed after (4.57) allow us to apply (4.49), with Ω = Sn−1 and with h0 and hε replaced by hε1,ε2 and hε1+ε,ε2, respectively, to obtain
∂
∂ε1VeG([hε1,ε2]) = lim
ε→0
VeG([hε1+ε,ε2]) − eVG([hε1,ε2])
ε = n
Z
Sn−1
g(u) d eCG,ψ([hε1,ε2], u), (4.64) and with g, h0, and hε replaced by 1, hε1,ε2 and hε1,ε2+ε, respectively, to yield
∂
∂ε2
VeG([hε1,ε2]) = n Z
Sn−1
1 d eCG,ψ([hε1,ε2], u) = n eCG,ψ([hε1,ε2], Sn−1) 6= 0. (4.65)
Since [hε1,ε2] depends continuously on ε1, ε2and in view of Proposition 4.3.1 (ii), (4.64) and (4.65) show that the gradient of the map (ε1, ε2) 7→ eVG([hε1,ε2]) has rank 1 and depends continuously on (ε1, ε2), implying that this map is continuously differentiable.
Hence we may apply the method of Lagrange multipliers to conclude from (4.62) that It follows from (4.66), (4.68), and (4.70) that
Z
and from (4.67), (4.69), and (4.71) that
τ = − |µ|
n eCG,ψ(K0, Sn−1). (4.73) In particular, we see from (4.73) that τ is independent of g. Finally, (4.72) and (4.73) show that (4.55) holds with K replaced by K0.
We remark that − eCG,ψ(K, ·) is a nonnegative measure since Gt < 0. Note that
(4.53) holds if limt→0+G(t, u) = ∞ for u ∈ Sn−1and limt→∞G(t, u) = 0 for u ∈ Σε(v).
This follows from the monotone convergence theorem, since t 7→ G(t, u) is decreasing.
In order to solve Problem 4.3.2 when t 7→ G(t, u) is increasing, one needs to use different techniques and we leave it for future work in Chapter 5.
When ψ ≡ 1 (and hence ϕ(t) = log t ∈ J−∞), the following result was proved in [69, Theorem 5.1].
Corollary 4.3.4. Let µ be a nonzero finite Borel measure on Sn−1 not concentrated on any closed hemisphere. Let φ : Rn\ {o} → (0, ∞) be continuous and such that Φ is continuous on (0, ∞) × Sn−1, where Φ is defined by (4.2). Let 0 < c < 1 and suppose that for v ∈ Sn−1,
b→0+lim Vφ(C(v, b, c)) = ∞, (4.74) where C(v, b, c) = {x ∈ Rn: |x| ≥ b and hx/|x|, vi ≥ c} and Vφ(·) is defined by (4.3).
Let ψ : (0, ∞) → (0, ∞) be continuous and satisfy (4.54). Then there exists K ∈ K(o)n such that
µ
|µ| = Ceφ,ψ(K, ·) Ceφ,ψ(K, Sn−1).
Proof. By assumption, Φ is continuous on (0, ∞) × Sn−1, and limt→∞Φ(t, u) = 0 for u ∈ Sn−1. Hence the second condition in (4.53) holds with G replaced by Φ. Clearly,
∂Φ(t, u)/∂t = −φ(tu)tn−1 < 0. By (4.74),
∞ = lim
b→0+Vφ(C(v, b, c)) = lim
b→0+
Z
Σc(v)
Z ∞ b
φ(ru)rn−1drdu = lim
b→0+
Z
Σc(v)
Φ(b, u)du.
Therefore the first condition in (4.53) also holds with G replaced by Φ. Due to the fact eCΦ,ψ(K, ·) = − eCφ,ψ(K, ·), Theorem 4.3.3 yields the result.
Another special case arises if µ is a discrete measure on Sn−1, that is, µ = Pm
i=1ciδvi, where ci > 0 for i = 1, . . . , m, and v1, . . . , vm ∈ Sn−1 are not contained in any closed hemisphere. Let G and ψ be as in Theorem 4.3.3. Then there exists a polytope P ∈ K(o)n such that
µ
|µ| = CeG,ψ(P, ·) CeG,ψ(P, Sn−1).
To see this, note that Theorem 4.3.3 ensures the existence of a K ∈ K(o)n such that (4.55) holds. Since µ is discrete, we obtain
CeG,ψ(K, ·) = Xm
i=1
¯ ciδvi,
where ¯ci = eCG,ψ(K, Sn−1)ci/|µ| < 0 for i = 1, . . . , m. Proposition 4.3.1 (i) shows that there is a measurable function g : Sn−1 → (−∞, 0] such that
Xm i=1
¯
ciδvi(E) = Z
E
g(u) dS(K, u)
for Borel sets E ⊂ Sn−1. Hence S(K, ·) is a discrete measure and [59, Theorem 4.5.4]
implies that K is a polytope.